Grade 11

Advanced

6 min

Lesson 1 of 12

Wave Fundamentals and Phase Relationships

Not Started

A deep dive into advanced wave properties, focusing on phase difference and path difference in preparation for interference studies.

Ever wondered how noise-canceling headphones turn a roaring jet engine into a whisper? It's not magic—it's the precise mathematical manipulation of wave timing known as phase.

Learning Objectives

1.Define phase difference in degrees, radians, and fractions of a cycle
2.Calculate the path difference ( $\Delta x$ ) between two coherent wave sources
3.Apply the mathematical relationship between path difference and phase difference

Understanding Phase Difference

In wave mechanics, phase describes the specific stage of a wave's cycle at a given point in time. Imagine a circle: one full rotation is $360^\circ$ or $2\pi$ radians. Similarly, one full wave cycle is $2\pi$ radians. Phase difference ( $\phi$ ) measures how much one wave is 'out of step' with another. If two waves reach their peaks at the exact same time, their phase difference is $0$ . If one is at a peak while the other is at a trough, they are $180^\circ$ (or $\pi$ radians) out of phase, often called being in anti-phase.

Quick Check

If two waves are shifted by exactly one-quarter of a cycle, what is their phase difference in radians?

A full cycle is

2\pi

. Divide that by 4.

Answer

$\frac{\pi}{2}$ radians

The Concept of Path Difference

When two waves travel from different sources (

S_1

and

S_2

) to meet at a single point (

P

), they often travel different distances. The path difference (

\Delta x

) is simply the absolute difference between these two distances. It is measured in meters (m). Mathematically, we express this as:

\Delta x = |S_1P - S_2P|

This physical 'extra distance' is the primary reason waves arrive at a point with different phases, leading to the interference patterns we see in light and sound.

1. A sound wave from Speaker A travels

5.0

m to reach a listener. 2. A sound wave from Speaker B travels

5.7

m to reach the same listener. 3. Calculate the path difference:

\Delta x = |5.7\text{ m} - 5.0\text{ m}| = 0.7\text{ m}

The Bridge: Linking Path and Phase

The most critical concept in wave interference is the relationship between physical distance (path difference) and angular timing (phase difference). Since one full wavelength (

\lambda

) corresponds to a phase shift of

2\pi

radians, we can use a ratio to find the phase difference for any path difference:

\phi = \frac{\Delta x}{\lambda} \times 2\pi

This formula tells us that the phase difference is the fraction of the wavelength represented by the path difference, scaled to a full circle (

2\pi

Two coherent light sources with a wavelength of

600

nm (

6.0 \times 10^{-7}

m) meet at a point. The path difference is

300

nm (

3.0 \times 10^{-7}

m). 1. Identify the variables:

\Delta x = 3.0 \times 10^{-7}

\lambda = 6.0 \times 10^{-7}

m. 2. Apply the formula:

\phi = \frac{3.0 \times 10^{-7}}{6.0 \times 10^{-7}} \times 2\pi

3. Simplify:

\phi = 0.5 \times 2\pi = \pi

radians. 4. Conclusion: The waves are in anti-phase.

Quick Check

If the path difference is exactly equal to one full wavelength ( $\Delta x = \lambda$ ), what is the phase difference in degrees?

Substitute

\lambda

for

\Delta x

in the ratio.

Answer

$360^\circ$ (which is effectively the same as $0^\circ$ or being in-phase)

A radio station broadcasts at a frequency of

100

MHz (wavelength

\lambda = 3.0

m). A receiver is

15.0

m from Tower A and

19.5

m from Tower B. 1. Calculate path difference:

\Delta x = |19.5 - 15.0| = 4.5

m. 2. Find the phase difference in radians:

\phi = \frac{4.5}{3.0} \times 2\pi = 1.5 \times 2\pi = 3\pi \text{ rad}

3. Determine the state: Since

3\pi

is an odd multiple of

\pi

, the waves are in anti-phase, resulting in destructive interference (a 'dead zone').

Key Takeaways

1Phase difference ( $\phi$ ) measures the angular shift between waves, usually in radians ( $2\pi$ per cycle).
2Path difference ( $\Delta x$ ) is the physical distance gap between two wave paths: $|S_1P - S_2P|$ .
3The two are linked by the ratio: $\phi = \frac{2\pi \Delta x}{\lambda}$ .
4When $\Delta x$ is a whole number of wavelengths ( $n\lambda$ ), waves are in-phase; when it is a half-number ( $(n+0.5)\lambda$ ), they are in anti-phase.

Quick Check

Multiple Choice

What is the phase difference (in radians) for two waves that are exactly half a cycle apart?

Multiple Choice

If $\lambda = 10$ cm and the path difference is $5$ cm, what is the phase difference?

True/False

True or False: If the path difference is $2\lambda$ , the waves will arrive at the point in-phase.

What's Next?

Review Tomorrow

In 24 hours, try to sketch two waves with a phase difference of $\pi$ and explain how their path difference relates to their wavelength.

Practice Activity

Find the frequency of your favorite radio station, calculate its wavelength using $v = f\lambda$ (where $v = 3 \times 10^8$ m/s), and determine what path difference would cause a 'dead spot' in reception.

Browse

Browse

Wave Fundamentals and Phase Relationships

Learning Objectives

Understanding Phase Difference

The Concept of Path Difference

The Bridge: Linking Path and Phase

Key Takeaways

Quick Check

What's Next?

Wave Fundamentals and Phase Relationships

Learning Objectives

Understanding Phase Difference

The Concept of Path Difference

The Bridge: Linking Path and Phase

Key Takeaways

Quick Check

What's Next?